备注:在 LeetCode 中的运行时间也不是特别慢。 Java 实现 importjava.util.Map;importjava.util.HashMap;importjava.util.List;importjava.util.ArrayList;classSolution{publicList<Integer>topKFrequent(int[] nums,intk){// 统计元素的频率Map<Integer, Integer> freqMap =newHashMap<>();for(intnum : nums...
【leetcode】347. Top K Frequent Elements 题目地址:https://leetcode.com/problems/top-k-frequent-elements/ 从一个数组中求解出现次数最多的k个元素,本质是top k问题,用堆排序解决。 关于堆排序,其时间复杂度在最好和最坏的场景下都是O(nlogn)。 一开始想定义一个结构体,包含元素和元素个数两个成员,后...
347. Top K Frequent ElementsMedium Topics Companies Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order. Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] Example 2: Input: nums = [1], k ...
Top K Frequent Elements - LeetCode https://leetcode.com/problems/top-k-frequent-elements/solution/ Python O(n) solution without sort, without heap, without quickselect - LeetCode Discuss https://leetcode.com/problems/top-k-frequent-elements/discuss/81697/Python-O(n)-solution-without-sort-with...
3. 把top k Map.Entry入列 4. 把top k Map.Entry.getKey() 放入ArrayList 中 publicclassSolution {publicList<Integer> topKFrequent(int[] nums,intk) {//1. iterator all the distinct elements and countMap<Integer, Integer> map =newHashMap<Integer, Integer>();for(inti = 0; i < nums.lengt...
You may assumekis always valid, 1 ≤k≤ number of unique elements. Your algorithm's time complexity must be better than O(nlogn), wherenis the array's size. 对于这种统计出现个数的题目,一般用map或unordered_map进行统计先,我开始的想法是通过统计之后,再用sort函数对pair的second进行排序输出,不过...
LeetCode 【347. Top K Frequent Elements】 Given a non-empty array of integers, return thekmost frequent elements. For example, Given[1,1,1,2,2,3]and k = 2, return[1,2]. 其实最简单的就是想到就是用一个小顶堆实现,如果堆中元素个数小于K,则插入元素,如果大于K,则和堆顶比较,如果大于...
LeetCode - 347. Top K Frequent Elements Given a non-empty array of integers, return thekmost frequent elements. Example 1: Input: nums =[1,1,1,2,2,3], k =2 Output:[1,2] Example 2: Input: nums =[1], k =1 Output:[1]...
class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: # 统计 nums 中每个数字出现的次数, # 时间复杂度为 O(n) ,空间复杂度为 O(n) num_to_cnt = Counter(nums) #将 num_to_cnt 中的数字及其出现次数收集到数组中, # 时间复杂度为 O(n) ,空间复杂度为 O(n)...
You may assumekis always valid, 1 ≤k≤ number of unique elements. Your algorithm's time complexitymust bebetter than O(nlogn), wherenis the array's size. 给一个非空整数数组,返回前k个频率最高的元素。k总是合理的,1 ≤k1 ≤独立元素的数量。要求算法时间复杂度必须优于O(n log n),n是数...